Class 12: Inverse Trigonometric Functions

$\displaystyle \textbf{Question 1: }\text{Evaluate the following:}$
$\displaystyle \text{(i) }\tan\left\{2\tan^{-1}\frac15-\frac{\pi}{4}\right\}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \tan\left\{2\tan^{-1}\frac15-\frac{\pi}{4}\right\}$
$\displaystyle =\tan\left\{2\tan^{-1}\frac15-\tan^{-1}1\right\}$
$\displaystyle =\tan\left\{\tan^{-1}\left(\frac{2\cdot\frac15}{1-\left(\frac15\right)^2}\right)-\tan^{-1}1\right\}$
$\displaystyle =\tan\left\{\tan^{-1}\left(\frac{\frac25}{\frac{24}{25}}\right)-\tan^{-1}1\right\}$
$\displaystyle =\tan\left\{\tan^{-1}\frac5{12}-\tan^{-1}1\right\}$
$\displaystyle =\tan\left\{\tan^{-1}\left(\frac{\frac5{12}-1}{1+\frac5{12}\cdot1}\right)\right\}$
$\displaystyle \qquad\left[\because\ \tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right),\ xy>-1\right]$
$\displaystyle =\tan\left\{\tan^{-1}\left(\frac{-\frac7{12}}{\frac{17}{12}}\right)\right\}$
$\displaystyle =\tan\left(\tan^{-1}\left(-\frac7{17}\right)\right)$
$\displaystyle =-\frac7{17}$

$\displaystyle \textbf{Question 1: }\text{Evaluate the following:}$
$\displaystyle \text{(ii) }\tan\left(\frac12\sin^{-1}\frac34\right)\qquad\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(ii)}$
$\displaystyle \text{Let }\frac12\sin^{-1}\frac34=x.$
$\displaystyle \Rightarrow2x=\sin^{-1}\frac34$
$\displaystyle \Rightarrow\sin2x=\frac34$
$\displaystyle \Rightarrow\cos2x=\frac{\sqrt7}{4}\qquad\left[\because 0<2x<\frac{\pi}{2}\right]$
$\displaystyle \therefore\tan x=\sqrt{\frac{1-\cos2x}{1+\cos2x}}$
$\displaystyle =\sqrt{\frac{1-\frac{\sqrt7}{4}}{1+\frac{\sqrt7}{4}}}$
$\displaystyle =\sqrt{\frac{4-\sqrt7}{4+\sqrt7}}$
$\displaystyle =\sqrt{\frac{(4-\sqrt7)^2}{16-7}}$
$\displaystyle =\sqrt{\frac{(4-\sqrt7)^2}{9}}$
$\displaystyle =\frac{4-\sqrt7}{3}$
$\displaystyle \therefore\tan\left(\frac12\sin^{-1}\frac34\right)=\frac{4-\sqrt7}{3}$

$\displaystyle \textbf{Question 1: }\text{Evaluate the following:}$
$\displaystyle \text{(iii) }\sin\left(\frac12\cos^{-1}\frac45\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iii)}$
$\displaystyle \sin\left(\frac12\cos^{-1}\frac45\right)$
$\displaystyle =\sin\left(\sin^{-1}\sqrt{\frac{1-\frac45}{2}}\right)$
$\displaystyle \qquad\left[\because\ 2\sin^{-1}\sqrt{\frac{1-x}{2}}=\cos^{-1}x,\ 0\le x\le1\right]$
$\displaystyle =\sqrt{\frac{\frac15}{2}}$
$\displaystyle =\sqrt{\frac1{10}}$
$\displaystyle =\frac1{\sqrt{10}}=\frac{\sqrt{10}}{10}$

$\displaystyle \textbf{Question 1: }\text{Evaluate the following:}$
$\displaystyle \text{(iv) }\sin\left(2\tan^{-1}\frac23\right)+\cos\left(\tan^{-1}\sqrt3\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iv)}$
$\displaystyle \sin\left(2\tan^{-1}\frac23\right)+\cos\left(\tan^{-1}\sqrt3\right)$
$\displaystyle =\sin\left[\sin^{-1}\left(\frac{2\cdot\frac23}{1+\left(\frac23\right)^2}\right)\right]+\cos\left[\cos^{-1}\left(\frac1{\sqrt{1+(\sqrt3)^2}}\right)\right]$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\sin^{-1}\left(\frac{2x}{1+x^2}\right),\ -1\leq x\leq1\right]$
$\displaystyle =\sin\left[\sin^{-1}\left(\frac{\frac43}{\frac{13}{9}}\right)\right]+\cos\left(\cos^{-1}\frac12\right)$
$\displaystyle =\sin\left(\sin^{-1}\frac{12}{13}\right)+\cos\left(\cos^{-1}\frac12\right)$
$\displaystyle =\frac{12}{13}+\frac12$
$\displaystyle =\frac{24+13}{26}$
$\displaystyle =\frac{37}{26}$

$\displaystyle \textbf{Question 2: }\text{Prove the following results:}$
$\displaystyle \text{(i) }2\sin^{-1}\frac35=\tan^{-1}\frac{24}{7}$
$\displaystyle \text{(ii) }\tan^{-1}\frac14+\tan^{-1}\frac29=\frac12\cos^{-1}\frac35=\frac12\sin^{-1}\frac45\qquad\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \text{LHS}=2\sin^{-1}\frac35$
$\displaystyle =2\tan^{-1}\left[\frac{\frac35}{\sqrt{1-\left(\frac35\right)^2}}\right]$
$\displaystyle \qquad\left[\because\ \sin^{-1}x=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right),\ 0\leq x<1\right]$
$\displaystyle =2\tan^{-1}\left(\frac{\frac35}{\frac45}\right)$
$\displaystyle =2\tan^{-1}\frac34$
$\displaystyle =\tan^{-1}\left[\frac{2\cdot\frac34}{1-\left(\frac34\right)^2}\right]$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^2}\right),\ -1<x<1\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac32}{\frac7{16}}\right)$
$\displaystyle =\tan^{-1}\frac{24}{7}$
$\displaystyle =\text{RHS}$
$\displaystyle \\$
$\displaystyle \text{(ii)}$
$\displaystyle \tan^{-1}\frac14+\tan^{-1}\frac29$
$\displaystyle =\tan^{-1}\left[\frac{\frac14+\frac29}{1-\frac14\cdot\frac29}\right]$
$\displaystyle \qquad\left[\because\ \frac14\cdot\frac29<1\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac{17}{36}}{\frac{34}{36}}\right)$
$\displaystyle =\tan^{-1}\frac12$
$\displaystyle =\frac12\cos^{-1}\left[\frac{1-\left(\frac12\right)^2}{1+\left(\frac12\right)^2}\right]$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right),\ x\geq0\right]$
$\displaystyle =\frac12\cos^{-1}\left(\frac{1-\frac14}{1+\frac14}\right)$
$\displaystyle =\frac12\cos^{-1}\left(\frac{\frac34}{\frac54}\right)$
$\displaystyle =\frac12\cos^{-1}\frac35$
$\displaystyle \text{Also,}$
$\displaystyle \tan^{-1}\frac12=\frac12\sin^{-1}\left[\frac{2\left(\frac12\right)}{1+\left(\frac12\right)^2}\right]$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\sin^{-1}\left(\frac{2x}{1+x^2}\right),\ -1\leq x\leq1\right]$
$\displaystyle =\frac12\sin^{-1}\left(\frac{1}{1+\frac14}\right)$
$\displaystyle =\frac12\sin^{-1}\left(\frac{1}{\frac54}\right)$
$\displaystyle =\frac12\sin^{-1}\frac45$
$\displaystyle \therefore\tan^{-1}\frac14+\tan^{-1}\frac29=\frac12\cos^{-1}\frac35=\frac12\sin^{-1}\frac45$
$\displaystyle \\$
$\displaystyle \textbf{Question 2: }\text{Prove the following results:}$
$\displaystyle \text{(iii) }\tan^{-1}\frac23=\frac12\tan^{-1}\frac{12}{5}$
$\displaystyle \text{(iv) }\tan^{-1}\frac17+2\tan^{-1}\frac13=\frac{\pi}{4}\qquad\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iii)}$
$\displaystyle \text{LHS}=\tan^{-1}\frac23$
$\displaystyle =\frac12\left[2\tan^{-1}\frac23\right]$
$\displaystyle =\frac12\tan^{-1}\left[\frac{2\cdot\frac23}{1-\left(\frac23\right)^2}\right]$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^2}\right),\ -1<x<1\right]$
$\displaystyle =\frac12\tan^{-1}\left(\frac{\frac43}{\frac59}\right)$
$\displaystyle =\frac12\tan^{-1}\frac{12}{5}$
$\displaystyle =\text{RHS}$
$\displaystyle \\$
$\displaystyle \text{(iv)}$
$\displaystyle \text{LHS}=\tan^{-1}\frac17+2\tan^{-1}\frac13$
$\displaystyle =\tan^{-1}\frac17+\tan^{-1}\left[\frac{2\cdot\frac13}{1-\left(\frac13\right)^2}\right]$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^2}\right),\ -1<x<1\right]$
$\displaystyle =\tan^{-1}\frac17+\tan^{-1}\frac34$
$\displaystyle =\tan^{-1}\left[\frac{\frac17+\frac34}{1-\frac17\cdot\frac34}\right]$
$\displaystyle \qquad\left[\because\ \frac17\cdot\frac34<1\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac{25}{28}}{\frac{25}{28}}\right)$
$\displaystyle =\tan^{-1}1$
$\displaystyle =\frac{\pi}{4}$
$\displaystyle =\text{RHS}$
$\displaystyle \\$
$\displaystyle \textbf{Question 2: }\text{Prove the following results:}$
$\displaystyle \text{(v) }\sin^{-1}\frac45+2\tan^{-1}\frac13=\frac{\pi}{2}$
$\displaystyle \text{(vi) }2\sin^{-1}\frac35-\tan^{-1}\frac{17}{31}=\frac{\pi}{4}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(v)}$
$\displaystyle \text{LHS}=\sin^{-1}\frac45+2\tan^{-1}\frac13$
$\displaystyle =\sin^{-1}\frac45+\tan^{-1}\left[\frac{2\cdot\frac13}{1-\left(\frac13\right)^2}\right]$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^2}\right),\ -1<x<1\right]$
$\displaystyle =\sin^{-1}\frac45+\tan^{-1}\left(\frac{\frac23}{\frac89}\right)$
$\displaystyle =\sin^{-1}\frac45+\tan^{-1}\frac34$
$\displaystyle =\sin^{-1}\frac45+\cos^{-1}\left(\frac{1}{\sqrt{1+\left(\frac34\right)^2}}\right)$
$\displaystyle \qquad\left[\because\ \tan^{-1}x=\cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right),\ x\geq0\right]$
$\displaystyle =\sin^{-1}\frac45+\cos^{-1}\frac45$
$\displaystyle =\frac{\pi}{2}$
$\displaystyle =\text{RHS}$
$\displaystyle \\$
$\displaystyle \text{(vi)}$
$\displaystyle \text{LHS}=2\sin^{-1}\frac35-\tan^{-1}\frac{17}{31}$
$\displaystyle =2\tan^{-1}\left[\frac{\frac35}{\sqrt{1-\left(\frac35\right)^2}}\right]-\tan^{-1}\frac{17}{31}$
$\displaystyle \qquad\left[\because\ \sin^{-1}x=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right),\ 0\leq x<1\right]$
$\displaystyle =2\tan^{-1}\left(\frac{\frac35}{\frac45}\right)-\tan^{-1}\frac{17}{31}$
$\displaystyle =2\tan^{-1}\frac34-\tan^{-1}\frac{17}{31}$
$\displaystyle =\tan^{-1}\left[\frac{2\cdot\frac34}{1-\left(\frac34\right)^2}\right]-\tan^{-1}\frac{17}{31}$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^2}\right),\ -1<x<1\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac32}{\frac7{16}}\right)-\tan^{-1}\frac{17}{31}$
$\displaystyle =\tan^{-1}\frac{24}{7}-\tan^{-1}\frac{17}{31}$
$\displaystyle =\tan^{-1}\left[\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7}\cdot\frac{17}{31}}\right]$
$\displaystyle \qquad\left[\because\ \frac{24}{7}\cdot\frac{17}{31}>-1\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right)$
$\displaystyle =\tan^{-1}1$
$\displaystyle =\frac{\pi}{4}$
$\displaystyle =\text{RHS}$
$\displaystyle \\$
$\displaystyle \textbf{Question 2: }\text{Prove the following results:}$
$\displaystyle \text{(vii) }2\tan^{-1}\frac15+\tan^{-1}\frac18=\tan^{-1}\frac47$
$\displaystyle \text{(viii) }2\tan^{-1}\frac34-\tan^{-1}\frac{17}{31}=\frac{\pi}{4}\qquad\text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(vii)}$
$\displaystyle \text{LHS}=2\tan^{-1}\frac15+\tan^{-1}\frac18$
$\displaystyle =\tan^{-1}\left[\frac{2\cdot\frac15}{1-\left(\frac15\right)^2}\right]+\tan^{-1}\frac18$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^2}\right),\ -1<x<1\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac25}{\frac{24}{25}}\right)+\tan^{-1}\frac18$
$\displaystyle =\tan^{-1}\frac5{12}+\tan^{-1}\frac18$
$\displaystyle =\tan^{-1}\left[\frac{\frac5{12}+\frac18}{1-\frac5{12}\cdot\frac18}\right]$
$\displaystyle \qquad\left[\because\ \frac5{12}\cdot\frac18<1\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac{13}{24}}{\frac{91}{96}}\right)$
$\displaystyle =\tan^{-1}\frac47$
$\displaystyle =\text{RHS}$
$\displaystyle \\$
$\displaystyle \text{(viii)}$
$\displaystyle \text{LHS}=2\tan^{-1}\frac34-\tan^{-1}\frac{17}{31}$
$\displaystyle =\tan^{-1}\left[\frac{2\cdot\frac34}{1-\left(\frac34\right)^2}\right]-\tan^{-1}\frac{17}{31}$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^2}\right),\ -1<x<1\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac32}{\frac7{16}}\right)-\tan^{-1}\frac{17}{31}$
$\displaystyle =\tan^{-1}\frac{24}{7}-\tan^{-1}\frac{17}{31}$
$\displaystyle =\tan^{-1}\left[\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7}\cdot\frac{17}{31}}\right]$
$\displaystyle \qquad\left[\because\ \frac{24}{7}\cdot\frac{17}{31}>-1\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right)$
$\displaystyle =\tan^{-1}1$
$\displaystyle =\frac{\pi}{4}$
$\displaystyle =\text{RHS}$
$\displaystyle \\$
$\displaystyle \textbf{Question 2: }\text{Prove the following results:}$
$\displaystyle \text{(ix) }2\tan^{-1}\frac12+\tan^{-1}\frac17=\tan^{-1}\frac{31}{17}$
$\displaystyle \text{(x) }4\tan^{-1}\frac15-\tan^{-1}\frac1{239}=\frac{\pi}{4}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(ix)}$
$\displaystyle \text{LHS}=2\tan^{-1}\frac12+\tan^{-1}\frac17$
$\displaystyle =\tan^{-1}\left[\frac{2\cdot\frac12}{1-\left(\frac12\right)^2}\right]+\tan^{-1}\frac17$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^2}\right),\ -1<x<1\right]$
$\displaystyle =\tan^{-1}\left(\frac{1}{\frac34}\right)+\tan^{-1}\frac17$
$\displaystyle =\tan^{-1}\frac43+\tan^{-1}\frac17$
$\displaystyle =\tan^{-1}\left[\frac{\frac43+\frac17}{1-\frac43\cdot\frac17}\right]$
$\displaystyle \qquad\left[\because\ \frac43\cdot\frac17<1\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right)$
$\displaystyle =\tan^{-1}\frac{31}{17}$
$\displaystyle =\text{RHS}$
$\displaystyle \\$
$\displaystyle \text{(x)}$
$\displaystyle \text{LHS}=4\tan^{-1}\frac15-\tan^{-1}\frac1{239}$
$\displaystyle =2\left(2\tan^{-1}\frac15\right)-\tan^{-1}\frac1{239}$
$\displaystyle =2\tan^{-1}\left[\frac{2\cdot\frac15}{1-\left(\frac15\right)^2}\right]-\tan^{-1}\frac1{239}$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^2}\right),\ -1<x<1\right]$
$\displaystyle =2\tan^{-1}\frac5{12}-\tan^{-1}\frac1{239}$
$\displaystyle =\tan^{-1}\left[\frac{2\cdot\frac5{12}}{1-\left(\frac5{12}\right)^2}\right]-\tan^{-1}\frac1{239}$
$\displaystyle =\tan^{-1}\left(\frac{\frac56}{\frac{119}{144}}\right)-\tan^{-1}\frac1{239}$
$\displaystyle =\tan^{-1}\frac{120}{119}-\tan^{-1}\frac1{239}$
$\displaystyle =\tan^{-1}\left[\frac{\frac{120}{119}-\frac1{239}}{1+\frac{120}{119}\cdot\frac1{239}}\right]$
$\displaystyle \qquad\left[\because\ \frac{120}{119}\cdot\frac1{239}>-1\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac{28561}{28441}}{\frac{28561}{28441}}\right)$
$\displaystyle =\tan^{-1}1$
$\displaystyle =\frac{\pi}{4}$
$\displaystyle =\text{RHS}$

$\displaystyle \textbf{Question 3: }\text{If }\sin^{-1}\frac{2a}{1+a^2}-\cos^{-1}\frac{1-b^2}{1+b^2}=\tan^{-1}\frac{2x}{1-x^2},\text{ then prove that} \\ x=\frac{a-b}{1+ab},\quad 0\leq a,b\leq1,\ -1\leq x\leq1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a=\tan m,\quad b=\tan n,\quad x=\tan y.$
$\displaystyle \text{Since }0\leq a,b\leq1,\text{ we have }0\leq m,n\leq\frac{\pi}{4}.$
$\displaystyle \text{Also, since }-1\leq x\leq1,\text{ we have }-\frac{\pi}{4}\leq y\leq\frac{\pi}{4}.$
$\displaystyle \sin^{-1}\frac{2a}{1+a^2}-\cos^{-1}\frac{1-b^2}{1+b^2}=\tan^{-1}\frac{2x}{1-x^2}$
$\displaystyle \Rightarrow\sin^{-1}\left(\frac{2\tan m}{1+\tan^2m}\right)-\cos^{-1}\left(\frac{1-\tan^2n}{1+\tan^2n}\right)=\tan^{-1}\left(\frac{2\tan y}{1-\tan^2y}\right)$
$\displaystyle \Rightarrow\sin^{-1}(\sin2m)-\cos^{-1}(\cos2n)=\tan^{-1}(\tan2y)$
$\displaystyle \Rightarrow2m-2n=2y$
$\displaystyle \qquad\left[\because\ 0\leq2m,2n\leq\frac{\pi}{2}\text{ and }-\frac{\pi}{2}\leq2y\leq\frac{\pi}{2}\right]$
$\displaystyle \Rightarrow m-n=y$
$\displaystyle \Rightarrow\tan y=\tan(m-n)$
$\displaystyle \Rightarrow x=\frac{\tan m-\tan n}{1+\tan m\tan n}$
$\displaystyle \Rightarrow x=\frac{a-b}{1+ab}$

$\displaystyle \textbf{Question 4: }\text{Prove that:}$
$\displaystyle \text{(i) }\tan^{-1}\left(\frac{1-x^2}{2x}\right)+\cot^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{\pi}{2},\quad x\neq0$
$\displaystyle \text{(ii) }\sin\left\{\tan^{-1}\left(\frac{1-x^2}{2x}\right)+\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}=1,\quad x>0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \text{LHS}=\tan^{-1}\left(\frac{1-x^2}{2x}\right)+\cot^{-1}\left(\frac{1-x^2}{2x}\right)$
$\displaystyle =\tan^{-1}\left(\frac{1-x^2}{2x}\right)+\frac{\pi}{2}-\tan^{-1}\left(\frac{1-x^2}{2x}\right)$
$\displaystyle \qquad\left[\because\ \cot^{-1}t=\frac{\pi}{2}-\tan^{-1}t\right]$
$\displaystyle =\frac{\pi}{2}$
$\displaystyle =\text{RHS}$

$\displaystyle \text{(ii)}$
$\displaystyle \text{LHS}=\sin\left\{\tan^{-1}\left(\frac{1-x^2}{2x}\right)+\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}$
$\displaystyle =\sin\left\{\sin^{-1}\left[\frac{\frac{1-x^2}{2x}}{\sqrt{1+\left(\frac{1-x^2}{2x}\right)^2}}\right]+\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}$
$\displaystyle \qquad\left[\because\ \tan^{-1}t=\sin^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)\right]$
$\displaystyle =\sin\left\{\sin^{-1}\left[\frac{\frac{1-x^2}{2x}}{\sqrt{\frac{4x^2+(1-x^2)^2}{4x^2}}}\right]+\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}$
$\displaystyle =\sin\left\{\sin^{-1}\left[\frac{\frac{1-x^2}{2x}}{\frac{1+x^2}{2x}}\right]+\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}$
$\displaystyle \qquad\left[\because\ x>0\right]$
$\displaystyle =\sin\left\{\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)+\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}$
$\displaystyle =\sin\frac{\pi}{2}$
$\displaystyle \qquad\left[\because\ \sin^{-1}t+\cos^{-1}t=\frac{\pi}{2},\ -1\leq t\leq1\right]$
$\displaystyle =1$
$\displaystyle =\text{RHS}$

$\displaystyle \textbf{Question 5: }\text{If }\sin^{-1}\frac{2a}{1+a^2}+\sin^{-1}\frac{2b}{1+b^2}=2\tan^{-1}x,\text{ then prove that} \\ x=\frac{a+b}{1-ab},\quad -1\leq a,b\leq1,\ ab<1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a=\tan z\text{ and }b=\tan y.$
$\displaystyle \text{Since }-1\leq a,b\leq1,\text{ we have }-\frac{\pi}{4}\leq z,y\leq\frac{\pi}{4}.$
$\displaystyle \sin^{-1}\frac{2a}{1+a^2}+\sin^{-1}\frac{2b}{1+b^2}=2\tan^{-1}x$
$\displaystyle \Rightarrow\sin^{-1}\left(\frac{2\tan z}{1+\tan^2z}\right)+\sin^{-1}\left(\frac{2\tan y}{1+\tan^2y}\right)=2\tan^{-1}x$
$\displaystyle \Rightarrow\sin^{-1}(\sin2z)+\sin^{-1}(\sin2y)=2\tan^{-1}x$
$\displaystyle \Rightarrow2z+2y=2\tan^{-1}x$
$\displaystyle \qquad\left[\because\ -\frac{\pi}{2}\leq2z,2y\leq\frac{\pi}{2}\right]$
$\displaystyle \Rightarrow z+y=\tan^{-1}x$
$\displaystyle \Rightarrow\tan^{-1}a+\tan^{-1}b=\tan^{-1}x$
$\displaystyle \Rightarrow\tan^{-1}\left(\frac{a+b}{1-ab}\right)=\tan^{-1}x$
$\displaystyle \qquad\left[\because\ ab<1\right]$
$\displaystyle \Rightarrow\frac{a+b}{1-ab}=x$
$\displaystyle \therefore x=\frac{a+b}{1-ab}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Show that }2\tan^{-1}x+\sin^{-1}\frac{2x}{1+x^2}\text{ is a constant for }x\geq1,\text{ and find that constant.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Consider }2\tan^{-1}x+\sin^{-1}\frac{2x}{1+x^2}.$
$\displaystyle \text{(i) For }x>1,$
$\displaystyle 2\tan^{-1}x+\sin^{-1}\frac{2x}{1+x^2}$
$\displaystyle =\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right)+\sin^{-1}\frac{2x}{1+x^2}$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right),\ x>1\right]$
$\displaystyle =\pi$
$\displaystyle \text{(ii) For }x=1,$
$\displaystyle 2\tan^{-1}x+\sin^{-1}\frac{2x}{1+x^2}$
$\displaystyle =2\tan^{-1}1+\sin^{-1}\frac{2\cdot1}{1+1^2}$
$\displaystyle =2\tan^{-1}1+\sin^{-1}1$
$\displaystyle =2\cdot\frac{\pi}{4}+\frac{\pi}{2}$
$\displaystyle =\frac{\pi}{2}+\frac{\pi}{2}$
$\displaystyle =\pi$
$\displaystyle \therefore 2\tan^{-1}x+\sin^{-1}\frac{2x}{1+x^2}\text{ is constant for }x\geq1.$
$\displaystyle \text{Hence, the required constant is }\pi.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Find the value of each of the following:}$
$\displaystyle \text{(i) }\tan^{-1}\left\{2\cos\left(2\sin^{-1}\frac12\right)\right\}$
$\displaystyle \text{(ii) }\cos\left(\sec^{-1}x+\mathrm{cosec}^{-1}x\right),\quad |x|\geq1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \text{Let }\sin^{-1}\frac12=y.$
$\displaystyle \Rightarrow\sin y=\frac12$
$\displaystyle \tan^{-1}\left\{2\cos\left(2\sin^{-1}\frac12\right)\right\}$
$\displaystyle =\tan^{-1}(2\cos2y)$
$\displaystyle =\tan^{-1}\left[2(1-2\sin^2y)\right]$
$\displaystyle \qquad\left[\because\ \cos2y=1-2\sin^2y\right]$
$\displaystyle =\tan^{-1}\left[2\left(1-2\cdot\frac14\right)\right]$
$\displaystyle =\tan^{-1}\left(2\cdot\frac12\right)$
$\displaystyle =\tan^{-1}1$
$\displaystyle =\frac{\pi}{4}$

$\displaystyle \text{(ii)}$
$\displaystyle \cos\left(\sec^{-1}x+\mathrm{cosec}^{-1}x\right)$
$\displaystyle =\cos\frac{\pi}{2}$
$\displaystyle \qquad\left[\because\ \sec^{-1}x+\mathrm{cosec}^{-1}x=\frac{\pi}{2},\ |x|\geq1\right]$
$\displaystyle =0$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Solve the following equations for }x:$
$\displaystyle \text{(i) }\tan^{-1}\frac14+2\tan^{-1}\frac15+\tan^{-1}\frac16+\tan^{-1}\frac1x=\frac{\pi}{4}$
$\displaystyle \text{(ii) }3\sin^{-1}\frac{2x}{1+x^2}-4\cos^{-1}\frac{1-x^2}{1+x^2}+2\tan^{-1}\frac{2x}{1-x^2}=\frac{\pi}{3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \tan^{-1}\frac14+2\tan^{-1}\frac15+\tan^{-1}\frac16+\tan^{-1}\frac1x=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan^{-1}\frac14+\tan^{-1}\left[\frac{2\cdot\frac15}{1-\left(\frac15\right)^2}\right]+\tan^{-1}\frac16+\tan^{-1}\frac1x=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan^{-1}\frac14+\tan^{-1}\frac5{12}+\tan^{-1}\frac16+\tan^{-1}\frac1x=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan^{-1}\left[\frac{\frac14+\frac5{12}}{1-\frac14\cdot\frac5{12}}\right]+\tan^{-1}\frac16+\tan^{-1}\frac1x=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan^{-1}\frac{32}{43}+\tan^{-1}\frac16+\tan^{-1}\frac1x=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan^{-1}\left[\frac{\frac{32}{43}+\frac16}{1-\frac{32}{43}\cdot\frac16}\right]+\tan^{-1}\frac1x=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan^{-1}\frac{235}{226}+\tan^{-1}\frac1x=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan^{-1}\frac1x=\frac{\pi}{4}-\tan^{-1}\frac{235}{226}$
$\displaystyle \Rightarrow\frac1x=\tan\left(\frac{\pi}{4}-\tan^{-1}\frac{235}{226}\right)$
$\displaystyle \Rightarrow\frac1x=\frac{1-\frac{235}{226}}{1+\frac{235}{226}}$
$\displaystyle \Rightarrow\frac1x=\frac{-\frac9{226}}{\frac{461}{226}}$
$\displaystyle \Rightarrow\frac1x=-\frac9{461}$
$\displaystyle \Rightarrow x=-\frac{461}{9}$
$\displaystyle \\$

$\displaystyle \text{(ii)}$
$\displaystyle \text{Let }\tan^{-1}x=\theta,\quad-\frac{\pi}{2}<\theta<\frac{\pi}{2}.$
$\displaystyle \therefore x=\tan\theta.$
$\displaystyle \text{Also, }x\neq\pm1.$
$\displaystyle \text{Case I: }0\leq x<1$
$\displaystyle \Rightarrow0\leq\theta<\frac{\pi}{4}$
$\displaystyle \sin^{-1}\frac{2x}{1+x^2}=2\theta$
$\displaystyle \cos^{-1}\frac{1-x^2}{1+x^2}=2\theta$
$\displaystyle \tan^{-1}\frac{2x}{1-x^2}=2\theta$
$\displaystyle \therefore3(2\theta)-4(2\theta)+2(2\theta)=\frac{\pi}{3}$
$\displaystyle \Rightarrow2\theta=\frac{\pi}{3}$
$\displaystyle \Rightarrow\theta=\frac{\pi}{6}$
$\displaystyle \Rightarrow x=\tan\frac{\pi}{6}=\frac1{\sqrt3}$
$\displaystyle \text{Since }0<\frac1{\sqrt3}<1,\text{ this solution is valid.}$
$\displaystyle \text{Case II: }-1<x<0$
$\displaystyle \Rightarrow-\frac{\pi}{4}<\theta<0$
$\displaystyle \sin^{-1}\frac{2x}{1+x^2}=2\theta$
$\displaystyle \cos^{-1}\frac{1-x^2}{1+x^2}=-2\theta$
$\displaystyle \tan^{-1}\frac{2x}{1-x^2}=2\theta$
$\displaystyle \therefore3(2\theta)-4(-2\theta)+2(2\theta)=\frac{\pi}{3}$
$\displaystyle \Rightarrow18\theta=\frac{\pi}{3}$
$\displaystyle \Rightarrow\theta=\frac{\pi}{54},\text{ which contradicts }\theta<0.$
$\displaystyle \therefore\text{ there is no solution in }-1<x<0.$
$\displaystyle \text{Case III: }x>1$
$\displaystyle \Rightarrow\frac{\pi}{4}<\theta<\frac{\pi}{2}$
$\displaystyle \sin^{-1}\frac{2x}{1+x^2}=\pi-2\theta$
$\displaystyle \cos^{-1}\frac{1-x^2}{1+x^2}=2\theta$
$\displaystyle \tan^{-1}\frac{2x}{1-x^2}=2\theta-\pi$
$\displaystyle \therefore3(\pi-2\theta)-4(2\theta)+2(2\theta-\pi)=\frac{\pi}{3}$
$\displaystyle \Rightarrow\pi-10\theta=\frac{\pi}{3}$
$\displaystyle \Rightarrow\theta=\frac{\pi}{15},\text{ which contradicts }\theta>\frac{\pi}{4}.$
$\displaystyle \therefore\text{ there is no solution for }x>1.$
$\displaystyle \text{Case IV: }x<-1$
$\displaystyle \Rightarrow-\frac{\pi}{2}<\theta<-\frac{\pi}{4}$
$\displaystyle \sin^{-1}\frac{2x}{1+x^2}=-\pi-2\theta$
$\displaystyle \cos^{-1}\frac{1-x^2}{1+x^2}=-2\theta$
$\displaystyle \tan^{-1}\frac{2x}{1-x^2}=2\theta+\pi$
$\displaystyle \therefore3(-\pi-2\theta)-4(-2\theta)+2(2\theta+\pi)=\frac{\pi}{3}$
$\displaystyle \Rightarrow-\pi+6\theta=\frac{\pi}{3}$
$\displaystyle \Rightarrow\theta=\frac{2\pi}{9},\text{ which contradicts }\theta<-\frac{\pi}{4}.$
$\displaystyle \therefore\text{ there is no solution for }x<-1.$
$\displaystyle \therefore x=\frac1{\sqrt3}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Prove that }2\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{\theta}{2}\right)=\cos^{-1}\left(\frac{a\cos\theta+b}{a+b\cos\theta}\right), \\ \text{where }a>b>0\text{ and }0\leq\theta<\pi.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=2\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{\theta}{2}\right)$
$\displaystyle =\cos^{-1}\left[\frac{1-\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{\theta}{2}\right)^2}{1+\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{\theta}{2}\right)^2}\right]$
$\displaystyle \qquad\left[\because\ 2\tan^{-1}x=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right),\ x\geq0\right]$
$\displaystyle =\cos^{-1}\left[\frac{1-\frac{a-b}{a+b}\tan^2\frac{\theta}{2}}{1+\frac{a-b}{a+b}\tan^2\frac{\theta}{2}}\right]$
$\displaystyle =\cos^{-1}\left[\frac{a+b-(a-b)\tan^2\frac{\theta}{2}}{a+b+(a-b)\tan^2\frac{\theta}{2}}\right]$
$\displaystyle =\cos^{-1}\left[\frac{a\left(1-\tan^2\frac{\theta}{2}\right)+b\left(1+\tan^2\frac{\theta}{2}\right)}{a\left(1+\tan^2\frac{\theta}{2}\right)+b\left(1-\tan^2\frac{\theta}{2}\right)}\right]$
$\displaystyle =\cos^{-1}\left[\frac{a\left(\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\right)+b}{a+b\left(\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\right)}\right]$
$\displaystyle \qquad\left[\text{Dividing the numerator and denominator by }1+\tan^2\frac{\theta}{2}\right]$
$\displaystyle =\cos^{-1}\left(\frac{a\cos\theta+b}{a+b\cos\theta}\right)$
$\displaystyle \qquad\left[\because\ \cos\theta=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\right]$
$\displaystyle =\text{RHS}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Prove that }\tan^{-1}\frac{2ab}{a^2-b^2}+\tan^{-1}\frac{2xy}{x^2-y^2}=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2}, \\ \text{where }\alpha=ax-by\text{ and }\beta=ay+bx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }u=\frac{2ab}{a^2-b^2}\text{ and }v=\frac{2xy}{x^2-y^2}.$
$\displaystyle \text{Assuming }uv<1,\text{ we have}$
$\displaystyle \tan^{-1}u+\tan^{-1}v=\tan^{-1}\left(\frac{u+v}{1-uv}\right).$
$\displaystyle \therefore\tan^{-1}\frac{2ab}{a^2-b^2}+\tan^{-1}\frac{2xy}{x^2-y^2}$
$\displaystyle =\tan^{-1}\left[\frac{\frac{2ab}{a^2-b^2}+\frac{2xy}{x^2-y^2}}{1-\frac{2ab}{a^2-b^2}\cdot\frac{2xy}{x^2-y^2}}\right]$
$\displaystyle =\tan^{-1}\left[\frac{2\left\{ab(x^2-y^2)+xy(a^2-b^2)\right\}}{(a^2-b^2)(x^2-y^2)-4abxy}\right]$
$\displaystyle =\tan^{-1}\left[\frac{2(abx^2-aby^2+a^2xy-b^2xy)}{a^2x^2-a^2y^2-b^2x^2+b^2y^2-4abxy}\right]$
$\displaystyle =\tan^{-1}\left[\frac{2(ax-by)(ay+bx)}{(ax-by)^2-(ay+bx)^2}\right]$
$\displaystyle \text{Since }\alpha=ax-by\text{ and }\beta=ay+bx,$
$\displaystyle =\tan^{-1}\left(\frac{2\alpha\beta}{\alpha^2-\beta^2}\right)$
$\displaystyle =\text{RHS}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{For }a,b,x,y>0,\text{ prove that} \\ \frac{2}{3}\tan^{-1}\left(\frac{3ab^2-a^3}{b^3-3a^2b}\right)+\frac{2}{3}\tan^{-1}\left(\frac{3xy^2-x^3}{y^3-3x^2y}\right) =\tan^{-1}\left(\frac{2\alpha\beta}{\alpha^2-\beta^2}\right), \\ \text{ where }\alpha=by-ax,\ \beta=bx+ay, \text{provided }\frac{a}{b}<\frac1{\sqrt3},\ \frac{x}{y}<\frac1{\sqrt3} \\ \text{ and }\tan^{-1}\frac{a}{b}+\tan^{-1}\frac{x}{y}<\frac{\pi}{4}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a=b\tan m\text{ and }x=y\tan n.$
$\displaystyle \therefore m=\tan^{-1}\frac{a}{b},\qquad n=\tan^{-1}\frac{x}{y}.$
$\displaystyle \text{Since }\frac{a}{b}<\frac1{\sqrt3}\text{ and }\frac{x}{y}<\frac1{\sqrt3},$
$\displaystyle 0<m,n<\frac{\pi}{6}.$
$\displaystyle \text{LHS}=\frac23\tan^{-1}\left(\frac{3ab^2-a^3}{b^3-3a^2b}\right)+\frac23\tan^{-1}\left(\frac{3xy^2-x^3}{y^3-3x^2y}\right)$
$\displaystyle =\frac23\tan^{-1}\left(\frac{3\tan m-\tan^3m}{1-3\tan^2m}\right)+\frac23\tan^{-1}\left(\frac{3\tan n-\tan^3n}{1-3\tan^2n}\right)$
$\displaystyle =\frac23\tan^{-1}(\tan3m)+\frac23\tan^{-1}(\tan3n)$
$\displaystyle \qquad\left[\because\ \tan3t=\frac{3\tan t-\tan^3t}{1-3\tan^2t}\right]$
$\displaystyle =\frac23(3m)+\frac23(3n)$
$\displaystyle \qquad\left[\because\ 0<3m,3n<\frac{\pi}{2}\right]$
$\displaystyle =2(m+n)$
$\displaystyle \text{Now,}$
$\displaystyle \tan(m+n)=\frac{\tan m+\tan n}{1-\tan m\tan n}$
$\displaystyle =\frac{\frac{a}{b}+\frac{x}{y}}{1-\frac{a}{b}\cdot\frac{x}{y}}$
$\displaystyle =\frac{ay+bx}{by-ax}$
$\displaystyle =\frac{\beta}{\alpha}.$
$\displaystyle \therefore\tan2(m+n)=\frac{2\tan(m+n)}{1-\tan^2(m+n)}$
$\displaystyle =\frac{2\frac{\beta}{\alpha}}{1-\frac{\beta^2}{\alpha^2}}$
$\displaystyle =\frac{2\alpha\beta}{\alpha^2-\beta^2}.$
$\displaystyle \text{Since }0<m+n<\frac{\pi}{4},\text{ we have }0<2(m+n)<\frac{\pi}{2}.$
$\displaystyle \therefore2(m+n)=\tan^{-1}\left(\frac{2\alpha\beta}{\alpha^2-\beta^2}\right)$
$\displaystyle \therefore\text{LHS}=\tan^{-1}\left(\frac{2\alpha\beta}{\alpha^2-\beta^2}\right)=\text{RHS}$
$\displaystyle \\$